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Math Help - Egyptian Fractions

  1. #1
    Forum Admin topsquark's Avatar
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    Egyptian Fractions

    This is related to a recent question (http://www.mathhelpforum.com/math-he...ead.php?t=2762) in the Homework Help section. What are "Egyptian fractions?"

    Thanks!

    -Dan
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  2. #2
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    Quote Originally Posted by topsquark
    This is related to a recent question (http://www.mathhelpforum.com/math-he...ead.php?t=2762) in the Homework Help section. What are "Egyptian fractions?"

    Thanks!

    -Dan
    Hello,

    the old egypts only used fractions with the numerator 1. To express other fractions there existed tables which show, how to assemble this fraction by those with numerator 1.

    An example: \frac{47}{60}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}

    I assume that those fractions with the numerator 1 are meant when you are talking about "Egyptian fractions".

    Have a look here: http://www.mcs.surrey.ac.uk/Personal.../egyptian.html

    Greetings

    EB
    Last edited by earboth; May 2nd 2006 at 10:41 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by topsquark
    This is related to a recent question (http://www.mathhelpforum.com/math-he...ead.php?t=2762) in the Homework Help section. What are "Egyptian fractions?"

    Thanks!

    -Dan
    The Ancient Egyptian method of dealing with fractions was to express them
    as a sum of unit fractions. For example:

    <br />
\frac{2}{3}=\frac{1}{2}+\frac{1}{6}<br />
.

    this can be written in list form [2,6], and it is usual not to repeat a denominator in the expansion.

    The representation of a fraction as EF is not unique since the identity:

    <br />
\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}<br />

    can always be used to extend a representation.

    That any fraction can be written as an EF can be shown using the Greedy
    algorithm (Fibonacci-Sylvester algorithm). This proceeds as follows:

    Code:
    assume: p, q positive integers such that p/q is a rational in lowest terms <1.
    
    let n be the largest integer such that 1/n<=p/q, and R=p/q-1/n
    EF=[]    //null list form Egyptian fraction
    
    Repeat until R==0
        EF=EF|n    //concatenate n to EF
        let n be the largest integer such that 1/n<=R, and R=R-1/n
    end
    
    assert: EF is an Egyptian fraction representation of p/q
    This can be shown to be guaranteed to terminate (I won't do that here) with
    EF then obviously an Egyptian Fraction representation of p/q.

    Further information can be found here


    RonL

    Note to self - must type faster in future
    Last edited by CaptainBlack; May 2nd 2006 at 01:49 PM.
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    hey CaptainBlack, what languages do you use to code? I know C++ now and am looking for another language to pick up, something in the direction of web design.

    This has nothing to do with math (kind of)
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by c_323_h
    hey CaptainBlack, what languages do you use to code? I know C++ now and am looking for another language to pick up, something in the direction of web design.

    This has nothing to do with math (kind of)
    That was pseudo-code made up for the occasion.

    Otherwise mainly Euler (a freeware Matlab like system), Matlab (and more
    Matlab like systems than you can shake a stick at), some C, and in the past
    FORTRAN (66 & 77), Pascal, Basic (of assorted varieties), SPSS, Coral 66.

    RonL
    Last edited by CaptainBlack; May 2nd 2006 at 01:19 PM.
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  6. #6
    Grand Panjandrum
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    There is a minor bug in the pseudo-code in my previous post which
    was deliberately put there to see if anyone was paying attention

    If we start with p=1, that is we already have a unit/Egyptian fraction we
    are at the mercy of how we interpret a "Repeat until" loop. If we test at
    the top of the loop we will not get the right result under these conditions.

    I think the following should correct the problem.

    Code:
    assume: p, q positive integers such that p/q is a rational in lowest terms <1.
    
    let n be the largest integer such that 1/n<=p/q, and R=p/q-1/n
    EF=[n]    //list form Egyptian fraction with one entry
    
    Repeat while R<>0
        let n be the largest integer such that 1/n<=R, and R=R-1/n
        EF=EF|n    //concatenate n to EF
    end
    
    assert: EF is an Egyptian fraction representation of p/q
    or:

    Code:
    assume: p, q positive integers such that p/q is a rational in lowest terms <1.
    
    let n be the largest integer such that 1/n<=p/q, and R=p/q-1/n
    EF=[]    //null list form Egyptian fraction
    
    
    if R==0 then
      EF=[n]
    else
      Repeat
          EF=EF|n    //concatenate n to EF
          let n be the largest integer such that 1/n<=R, and R=R-1/n
      Until R==0
    endif
    
    assert: EF is an Egyptian fraction representation of p/q
    RonL
    Last edited by CaptainBlack; May 4th 2006 at 03:04 AM.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack
    There is a minor bug in the pseudo-code in my previous post which
    was deliberately put there to see if anyone was paying attention

    If we start with p=1, that is we already have a unit/Egyptian fraction we
    are at the mercy of how we interpret a "Repeat until" loop. If we test at
    the top of the loop we will not get the right result under these conditions.

    I think the following should correct the problem.

    Code:
    assume: p, q positive integers such that p/q is a rational in lowest terms <1.
    
    let n be the largest integer such that 1/n<=p/q, and R=p/q-1/n
    EF=[n]    //list form Egyptian fraction with one entry
    
    Repeat while R<>0
        let n be the largest integer such that 1/n<=R, and R=R-1/n
        EF=EF|n    //concatenate n to EF
    end
    
    assert: EF is an Egyptian fraction representation of p/q
    RonL
    I'm bilingual in English and Calculus. I don't read codes. Coeds on the other hand...

    -Dan
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  8. #8
    Grand Panjandrum
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    Just to see who has been following this so far:

    Find seven distinct positive integers n_1,\ n_2,\ n_3,\ n_4,\ n_5\ n_6,\ n_7
    the recipricals of which sum to 1, that is, find:

    <br />
n_1,\ \dots\ n_7 \ \in \mathbb{N}<br />
,

    such that:

    <br />
\forall\ i,j \in \{1,\dots, 7\},\   i \ne j \Rightarrow n_i \ne n_j <br />
,

    and

    <br />
1=\frac{1}{n_1}+\ \dots\ +\frac{1}{n_7}<br />

    Extra:

    Give an algorithm to generate such a set of integers of any length greater
    than 2.

    RonL
    Last edited by CaptainBlack; May 7th 2006 at 02:36 AM.
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